Question

Find where f(x)=4x-tanx,-π/2 < x < π/2 is increasing or decreasing and find its maximum and minimum values.

Answer 1

function , f(x) = 4x - tanx , -π/2 < x < π/2

differentiating f(x) with respect to x,

f'(x) = 4 - sec²x

now, f'(x) = 4 - sec²x = 0

⇒4 - sec²x = 0

⇒sec²x = (2)²

⇒secx = ±2

⇒x = ±π/3

so, there are three intervals ; (-π/2, -π/3), (-π/3,π/3), (π/3, π/2)

let's check f'(x) > 0 in which intervals.

-π/3 < x < π/3 ⇒secx < 2

so, f'(x) = 4- sec²x > 0

hence, f(x) is increasing in (-π/3, π/3).

similarly, check f'(x) value in (-π/2, -π/3) and (π/3, π/2).

you will get, f'(x) < 0 in these intervals.

so, f(x) is decreasing in (-π/2, -π/3) U (π/3, π/2).

again differentiate with respect to x,

f"(x) = 0 - 2sec²x. tanx = -2sec²x.tanx

at x = π/3, f"(π/3) = -2sec²(π/3).tan(π/3) < 0

hence, f(x) is maximum at x = π/3

and maximum value of f(x) = f(π/3) = 4(π/3) - tan(π/3) = 4π/3 - √3

at x = -π/3, f"(x) = -2sec²(-π/3).tan(-π/3) > 0

hence, f(x) is minimum at x = -π/3

and minimum value of f(x) = f(-π/3) = 4(-π/3) - tan(-π/3) = -4π/3 + √3